UVA-10118 Free Candies Memory Search Classic

Source: Internet
Author: User

Thought:D[a][b][c][d] said from the first basket took a sugar, the second took a B sugar, the third took a C sugar, the fourth to take a D sugar at most can also get how much candy. First of all understand a problem: if you can take a,b,c,d, regardless of how to take, the last basket of the remaining candy color and number are the same. So once you have searched for a state that has been searched, you can return directly and no further search is necessary.

AC Code:

#include <cstdio> #include <algorithm> #include <cstring> #include <utility> #include < string> #include <iostream> #include <map> #include <set> #include <vector> #include <
Queue> #include <stack> using namespace std;
#define EPS 1e-10 #define INF 0x3f3f3f3f #define PI pair<int, int> const int MAXN = 40 + 2;
int CAND[4][MAXN], n;
int top[4], bit[30];
	void Init () {bit[0] = 1;
for (int i = 1; i <; ++i) bit[i] = bit[i-1] * 2;
	} int dfs (int color, int cnt) {if (d[top[0]][top[1]][top[2]][top[3]]! = 1) return d[top[0]][top[1]][top[2]][top[3]];
	if (cnt = = 5) return d[top[0]][top[1]][top[2]][top[3]] = 0;
	int ans = 0;
		for (int i = 0; i < 4; ++i) {if (Top[i] >= N) continue;
		int col = ++top[i];
		col = Bit[cand[i][col]];
			 if (col & color) {//basket already has the right color int a = 1 + DFS (Color-col, cnt-1);
		ans = max (ans, a);
			} else {int a = DFS (color + col, cnt + 1); Ans =Max (ans, a);
	} top[i]--;
} return d[top[0]][top[1]][top[2]][top[3]] = ans;
	} int main () {init ();
		while (scanf ("%d", &n) = = 1 && N) {memset (d,-1, sizeof (d));
		memset (top, 0, sizeof (top)); 
		for (int i = 1; I <= n; ++i) for (int j = 0; j < 4; ++j) scanf ("%d", &cand[j][i]);
	printf ("%d\n", DFS (0, 0));
} return 0; }

If there are any irregularities, please note.

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